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evablogger [386]
3 years ago
12

A roller coaster car has a mass of 400kg and a speed of 15m/s What will be the P.E of this roller coaster car at its highest poi

nt ,where KE =0 at that point
Physics
1 answer:
Nutka1998 [239]3 years ago
5 0

Answer:

3000J

Explanation:

Given parameters:

mass  = 400kg

speed  = 15m/s

Unknown:

P.E at the highest point of the roller coaster = ?

Solution:

Due to the law of conservation of energy, the potential energy at the highest point can be solved using the formula of the kinetic energy

  Potential energy  = \frac{1}{2} mv²

m is the mass

v is the velocity

    Potential energy =  \frac{1}{2}  x 400 x 15  = 3000J

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What is the velocity of an object that has a momentum of 4,000 kg-m/s and a mass of 115 kg? Round to the nearest hundredth.
julia-pushkina [17]
<h2>Answer: 34.78 m/s</h2>

Explanation:

The momentum p is given by the following equation:

p=m.V   (1)

Where:

m is the mass of the object

V is the velocity of the object

Finding the velocity from (1):

V=\frac{p}{m}   (2)

V=\frac{4000kg.m/s}{115kg}  

<u>Finally:</u>

V=34.78m/s >>>This is the velocity of the object

4 0
3 years ago
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A 2 kg object traveling at 5 m s on a frictionless horizontal surface collides head-on with and sticks to a 3 kg object initiall
Svetlanka [38]

Answer: (d)

Explanation:

Given

Mass of object m=2\ kg

Speed of object u=5\ m/s

Mass of object at rest M=3\ kg

Suppose after collision, speed is v

conserving momentum

\Rightarrow mu+0=(m+M)v\\\\\Rightarrow v=\dfrac{2\times 5}{2+3}\\\\\Rightarrow v=2\ m/s

Initial kinetic energy

k_1=\dfrac{1}{2}\times 2\times 5^2\\\\k_1=25\ J

Final kinetic energy

k_2=\dfrac{1}{2}\times (2+3)\times 2^2\\\\k_2=10\ J

So, it is clear there is decrease in kinetic energy . Thus, energy decreases and velocity becomes 2 m/s.

4 0
3 years ago
If you throw a baseball while riding a bike, the ball's speed relative to the ground is the sum of the bicycle's speed and the s
Ostrovityanka [42]
Take for example driving by with a cake in your hand, then dropping it while going 30 mph. It will not drop directly down, it will gradually go in the direction you were driving while falling.
This is true I believe, if I'm interpreting correctly.
5 0
3 years ago
In a game of pool, a cue ball rolls without slipping toward the stationary eight ball with a momentum of 0.23 kg. After the two
IrinaVladis [17]

This question involves the concepts of the law of conservation of momentum.

The magnitude of the final momentum of the eight ball is "0.22 N.s".

According to the law of conservation of momentum:

P_{i1}+P_{i2}=P_{f1}+P_{f2}

where,

P_{i1} = initial momentum of the cue ball = 0.23 N.s

P_{i2} = initial momentum of the eight ball = 0 N.s (since ball is initially at rest)

P_{f1} = final momentum of the cue ball = 0.01 N.s

P_{f2} = final momentum of the eight ball = ?

Therefore,

0.23\ N.s + 0\N.s = 0.01\ N.s+P_{f2}\\\\P_{f2} = 0.22\ N.s

Learn more about the law of conservation of momentum here:

brainly.com/question/1113396?referrer=searchResults

3 0
3 years ago
100 points! please help!!!
IRINA_888 [86]

Answer:

Explanation:

A plane flies due north (90° from east) with a velocity of 100 km/h for 2 hours.

With no wind, it will be 100*2 = 200 km north of its starting point.

But a steady wind blows southeast at 30 km/h at an angle of 315° from due east.

So the wind itself will blow the plane 30*2 = 60km at an angle of 315° from due east.

That is the same as 60*cos315° = 42.43km due east and 60*sin315° = -42.43km north.

Combining, the plane is at 42.43km due east and 200-42.43 = 157.57km due north from its starting point.

7 0
3 years ago
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